# Simple Velocity question

## Homework Statement

The instantaneous velocity of a parrticle is given by the expression:
$$v=u sin \omega t$$
The time of complete cycle is $$T= \frac{2 \pi}{\omega}$$

Find out the average speed!

## The Attempt at a Solution

After integration;
$$\frac{2u} { \pi}$$

tell me if I am wrong I will show my working if I am wrong.
Thanks a lot!

Hi ritwik

The average speed over the time of a complete cycle? I think you made a mistake. Look at the sin function again and think of it's average value. It can be done without integrating or any calculus whatsoever. Or am I wrong?

Hi ritwik

The average speed over the time of a complete cycle? I think you made a mistake. Look at the sin function again and think of it's average value. It can be done without integrating or any calculus whatsoever. Or am I wrong?

Lol!! Thats so clever armis. But whatever your method might be, the answer will be same. So what did u get?
Dont help people unless u get the answer urself. It might confuse the beginners like me all the more.

alphysicist
Homework Helper
Hi ritwik06,

Your result looks right to me.

what limits did you take for integrating?

Lol!! Thats so clever armis
Well, I only wanted to help and solve the problem in the process. So were is no reason to laugh at me

I just thought that if velocity varies from u to -u then the average value will be just zero. Of course if we are talking about the absolute value of u then $$\frac{2u}{\pi}$$ is the answer

Dont help people unless u get the answer urself. It might confuse the beginners like me all the more.

Sorry, I didn't mean that

alphysicist
Homework Helper
Hi jablonsky27,

what limits did you take for integrating?

I got the same answer as ritwik. Are you integrating the correct function (they ask for the average speed)?

hi alphysicist,
you dont even have to integrate the given function. the average value of a sine(or cosine) over one cycle is always zero. the velocity function mentioned, is a sine with no additional constant term. so its average value too over one cycle will be zero.
if, you really want to calculate it,

Vaverage = (1/T).$$\int$$(u.sinwt) over the limits 0 to T,

= (u/T).[-cos(w.2/$$\Pi$$T) + cos(0)]

= (u/T).[-1 + 1]

= 0

hopefully, it makes sense. first time with Latex.

hi alphysicist,
you dont even have to integrate the given function. the average value of a sine(or cosine) over one cycle is always zero. the velocity function mentioned, is a sine with no additional constant term. so its average value too over one cycle will be zero.
if, you really want to calculate it,

Vaverage = (1/T).$$\int$$(u.sinwt) over the limits 0 to T,

= (u/T).[-cos(w.2/$$\Pi$$T) + cos(0)]

= (u/T).[-1 + 1]

= 0

hopefully, it makes sense. first time with Latex.
Well if you go back and read the question, its written in bold face:
average speed

Well, I only wanted to help and solve the problem in the process. So were is no reason to laugh at me

I just thought that if velocity varies from u to -u then the average value will be just zero. Of course if we are talking about the absolute value of u then $$\frac{2u}{\pi}$$ is the answer

Sorry, I didn't mean that

Hey, I wasnt laughing at you dear. I was just wonering how u could take that out without integration. I mean the average speed. Anyways, thanks a lot for the effort.

alphysicist
Homework Helper
jablonsky,

hi alphysicist,
you dont even have to integrate the given function. the average value of a sine(or cosine) over one cycle is always zero. the velocity function mentioned, is a sine with no additional constant term. so its average value too over one cycle will be zero.
if, you really want to calculate it,

Vaverage = (1/T).$$\int$$(u.sinwt) over the limits 0 to T,

= (u/T).[-cos(w.2/$$\Pi$$T) + cos(0)]

= (u/T).[-1 + 1]

= 0

hopefully, it makes sense. first time with Latex.

I see ritwik already remarked about how we are looking for the average speed (not average velocity). So the integral to calculate would be:

$$\mbox{average speed }= \frac{1}{T}\ \int\limits_0^Tdt\ |u \sin(\omega t)|$$

Or you could restrict the integral to just one-fourth of a period.

oh ya. sorry. should learn to read questions more carefully in the future.